A light source of wavelength `lambda` illuminates a metal and ejects photoelectron with `(KE)^(max) = 1 eV.`
Another light source of wave length `(lambda)/(3),` ejects
photoelectrons from same metal with `(KE)^(max)=5 eV.`
Find the value of work function (eV) of metal.

Options

A. 1 eV
B. 2 eV
C. `0.5` eV
D. None of these

Answer

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident photons to the work function of the metal and the kinetic energy of the emitted photoelectrons.

### Step-by-step Solution:

1. **Understand the Photoelectric Effect**:
According to Einstein's photoelectric equation:
\[ E = \phi + KE_{max} \]
where \( E \) is the energy of the incident photon, \( \phi \) is the work function of the metal, and \( KE_{max} \) is the maximum kinetic energy of the emitted photoelectron.

2. **Calculate Energy for Wavelength \( \lambda \)**:
The energy of a photon can be expressed in terms of its wavelength \( \lambda \):
\[ E = \frac{hc}{\lambda} \]
For the first light source with wavelength \( \lambda \) and \( KE_{max} = 1 \, \text{eV} \):
\[ \frac{hc}{\lambda} = \phi + 1 \, \text{eV} \tag{1} \]

3. **Calculate Energy for Wavelength \( \frac{\lambda}{3} \)**:
For the second light source with wavelength \( \frac{\lambda}{3} \) and \( KE_{max} = 5 \, \text{eV} \):
\[ E = \frac{hc}{\frac{\lambda}{3}} = \frac{3hc}{\lambda} \]
Thus, we can write:
\[ \frac{3hc}{\lambda} = \phi + 5 \, \text{eV} \tag{2} \]

4. **Set Up the Equations**:
We now have two equations:
- From equation (1):
\[ \phi = \frac{hc}{\lambda} - 1 \tag{3} \]
- From equation (2):
\[ \phi = \frac{3hc}{\lambda} - 5 \tag{4} \]

5. **Equate the Two Expressions for Work Function**:
Set equation (3) equal to equation (4):
\[ \frac{hc}{\lambda} - 1 = \frac{3hc}{\lambda} - 5 \]

6. **Solve for \( \phi \)**:
Rearranging gives:
\[ -1 + 5 = \frac{3hc}{\lambda} - \frac{hc}{\lambda} \]
\[ 4 = \frac{2hc}{\lambda} \]
Thus, we can express \( \phi \):
\[ \phi = \frac{hc}{\lambda} - 1 = 2 \, \text{eV} \]

7. **Final Calculation**:
Substitute back to find \( \phi \):
\[ \phi = 2 \, \text{eV} \]

### Conclusion:
The work function \( \phi \) of the metal is \( 2 \, \text{eV} \).